Example 3.2. A winding drum xaises a cage of mass 500 kg through a height of 100 metres. The mass of the winding drum is 250 kg and has an effective radius of 0.5 m and radius of gyration is 0.35 m . The mass of the rope is 3kg/m The cage has, at first.an acceleration of 1.5m/s^2 until a velocity of 10m/s is reached, after which the velocity is constant until the cage nears the top and the final retardation is 6m/s^2 Find 1. The time taken for the cage to reach the top, 2 . The torque which must be applied to the drum at starting; and 3. The power at the end of acceleration period. Solution. Given : Vert _(C)=500kg;s=100m;m_(D)=250kg;r=0.5m;k=0.35m m=3kg/m

1. To find the time taken for the cage to reach the top, we can use the kinematic equation:<br /><br />\[ s = ut + \frac{1}{2}at^2 \]<br /><br />where:<br />- \( s \) is the distance (100 m)<br />- \( u \) is the initial velocity (0 m/s)<br />- \( a \) is the acceleration (1.5 m/s²)<br />- \( t \) is the time<br /><br />Plugging in the values, we get:<br /><br />\[ 100 = 0 + \frac{1}{2}1 \times t^2 \]<br /><br />\[ 100 = 0.75t^2 \]<br /><br />\[ t^2 = \frac{100}{0.75} \]<br /><br />\[ t^2 = 133.33 \]<br /><br />\[ t = \sqrt{133.33} \]<br /><br />\[ t \approx 11.55 \text{ seconds} \]<br /><br />So, the time taken for the cage to reach the top is approximately 11.55 seconds.<br /><br />2. To find the torque which must be applied to the drum at starting, we can use the formula for torque:<br /><br />\[ \tau = I \alpha \]<br /><br />where:<br />- \( \tau \) is the torque<br />- \( I \) is the moment of inertia<br />- \( \alpha \) is the angular acceleration<br /><br />The moment of inertia for a solid disk is given by:<br /><br />\[ I = \frac{1}{2} m r^2 \]<br /><br />where:<br />- \( m \) is the mass of the drum (250 kg)<br />- \( r \) is the radius of the drum (0.5 m)<br /><br />Plugging in the values, we get:<br /><br />\[ I = \frac{1}{2} \times 250 \times (0.5)^2 \]<br /><br />\[ I = 62.5 \text{ kg m}^2 \]<br /><br />Now, we can find the angular acceleration using the formula:<br /><br />\[ \alpha = \frac{a}{r} \]<br /><br />where:<br />- \( a \) is the acceleration (1.5 m/s²)<br />- \( r \) is the radius of the drum (0.5 m)<br /><br />Plugging in the values, we get:<br /><br />\[ \alpha = \frac{1.5}{0.5} \]<br /><br />\[ \alpha = 3 \text{ rad/s}^2 \]<br /><br />Now, we can find the torque:<br /><br />\[ \tau = 62.5 \times 3 \]<br /><br />\[ \tau = 187.5 \text{ N m} \]<br /><br />So, the torque which must be applied to the drum at starting is 187.5 N m.<br /><br />3. To find the power at the end of the acceleration period, we can use the formula for power:<br /><br />\[ P = Fv \]<br /><br />where:<br />- \( P \) is the power<br />- \( F \) is the force<br />- \( v \) is the velocity<br /><br />The force can be calculated using Newton's second law:<br /><br />\[ F = ma \]<br /><br />where:<br />- \( m \) is the mass of the cage (500 kg)<br />- \( a \) is the acceleration (1.5 m/s²)<br /><br />Plugging in the values, we get:<br /><br />\[ F = 500 \times 1.5 \]<br /><br />\[ F = 750 \text{ N} \]<br /><br />Now, we can find the power:<br /><br />\[ P = 750 \times 10 \]<br /><br />\[ P = 7500 \text{ W} \]<br /><br />So, the power at the end of the acceleration period is 7500 W.